What I want to prove is that if I have a sequence of random variable $Y_n$ uniformly integrable that converge to Y, so $Y_n\to Y$ then $E[Y_n]\to E[Y]$.
I'm looking at the following proof (after having proved that Y is integrable):
Let $Y_n^\theta=Y_n\mathbb 1_{|Y_n|<\theta}$ and let $Y^\theta=Y\mathbb 1_{|Y|<\theta}$
For all $\theta \in \mathbb R^+$ s.t. $P(|Y|=\theta)=0$ we have $Y_n^\theta \to Y^\theta$ a.s.and $E[Y_n^\theta] \to E[Y^\theta]$
$1)$ why this step follows? and why they hold on $P(|Y|=\theta)=0$?.
Continuing with the proof we thus have: $\limsup |E[Y_n-Y]| \le \limsup \big(|E[Y_n-Y_n^\theta]|+|E[Y_n^\theta-Y^\theta]|+|E[Y^\theta-Y]|\big)$
$2)$ Why this inequality hold?
Going on with the proof I read: since $P(|Y|=\theta)>0$ can be only valid for countably many $\theta$ I can find an increasing sequence $\theta_k$ with $P(|Y|=\theta_k)=0$ and $\theta_k \to \infty.$
$3)$ why $P(|Y|=\theta)>0$ can be only valid for countably many $\theta$? and why I need $P(|Y|=\theta_k)=0$ (this is connected to question $1$)?
I would appreciate a help.